Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $x = \dfrac{z^2 + 5z - 24}{6z + 48} \times \dfrac{-4z + 36}{z - 3} $
Solution: First factor the quadratic. $x = \dfrac{(z - 3)(z + 8)}{6z + 48} \times \dfrac{-4z + 36}{z - 3} $ Then factor out any other terms. $x = \dfrac{(z - 3)(z + 8)}{6(z + 8)} \times \dfrac{-4(z - 9)}{z - 3} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (z - 3)(z + 8) \times -4(z - 9) } { 6(z + 8) \times (z - 3) } $ $x = \dfrac{ -4(z - 3)(z + 8)(z - 9)}{ 6(z + 8)(z - 3)} $ Notice that $(z + 8)$ and $(z - 3)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ -4\cancel{(z - 3)}(z + 8)(z - 9)}{ 6(z + 8)\cancel{(z - 3)}} $ We are dividing by $z - 3$ , so $z - 3 \neq 0$ Therefore, $z \neq 3$ $x = \dfrac{ -4\cancel{(z - 3)}\cancel{(z + 8)}(z - 9)}{ 6\cancel{(z + 8)}\cancel{(z - 3)}} $ We are dividing by $z + 8$ , so $z + 8 \neq 0$ Therefore, $z \neq -8$ $x = \dfrac{-4(z - 9)}{6} $ $x = \dfrac{-2(z - 9)}{3} ; \space z \neq 3 ; \space z \neq -8 $